To locate the discontinuities of the function $f(x) = \ln(\tan^2(x))$, we need to identify when the argument of the logarithmic function is undefined or non-positive, as the logarithmic function is only defined for positive real numbers.

Step-by-Step Process:

1. Domain of the logarithmic function: The logarithmic function $\ln(y)$ is defined when $y > 0$. Therefore, for $f(x) = \ln(\tan^2(x))$, we need: $\tan^2(x) > 0$ Since $\tan^2(x)$ is always non-negative and is only zero when $\tan(x) = 0$, we must exclude points where $\tan(x) = 0$. This happens when: $x = n\pi \quad \text{for integers} \ n$ Thus, $x = n\pi$ are points where the logarithm is undefined due to $\tan^2(x) = 0$.

2. Discontinuities of the tangent function: The tangent function $\tan(x)$ has vertical asymptotes where $\tan(x)$ is undefined, which occurs when: $x = \frac{(2n+1)\pi}{2} \quad \text{for integers} \ n$ These points are where $\tan(x)$ becomes infinite, hence $\tan^2(x)$ is also undefined, leading to discontinuities.

Conclusion:

The function $f(x) = \ln(\tan^2(x))$ has discontinuities at the following points:

1. At $x = n\pi$ (where $\tan(x) = 0$, and the logarithm is undefined).
2. At $x = \frac{(2n+1)\pi}{2}$ (where $\tan(x)$ has vertical asymptotes).

These are the points where the function is either undefined or has a discontinuity due to the nature of $\tan(x)$ and the logarithmic function.

Would you like more details on this process or have any questions?

---

Here are 5 related questions:

1. What is the domain of the function $f(x) = \ln(\sin(x))$?
2. How do vertical asymptotes influence the behavior of a function?
3. Can $f(x) = \tan(x)$ ever have a horizontal asymptote?
4. How would you find the domain of a composition of two functions?
5. What is the derivative of $f(x) = \ln(\tan^2(x))$?

Tip: When dealing with logarithmic functions, always check the argument’s domain first to avoid undefined regions.